Prove that there is at least $1$ solution and finitely many solutions. 
Fix an integer $n$ such that $n \geq 2$. Consider the system of equations
  $$\begin{cases}
a+b-c=n \\ 
a^2+b^2-c^2=n
\end{cases}$$
  in three variables $a,b,c\in\mathbb{Z}$.
  Prove that it has at least $1$ solution and finitely many solutions.

For example, for $n = 2$, it seems that the only solutions are $\left(1,1,0\right)$ and $\left(3,3,4\right)$, as found by the following python code:
def find(n, N):
    for a in range(-N, N+1):
        for b in range(-N, N+1):
            c = a + b - n
            if a**2 + b**2 - c**2 == n:
                print a, b, c

find(2, 25)

This question is probably one of the hardest I have attempted yet. I have tried to solve it but couldn't. Is there any way I could get help here?
 A: For even $n$,
$$a=2n-1, b=\frac{3n}2, c=\frac{5n}2-1$$
is a solution, while for odd $n$,
$$a=2n, \space b=\frac{3n-1}2, \space c=\frac{5n-1}2$$
is one.
To prove that there are only finitely many solutions for a given $n\ge2$, we observe that
$$b-c=n-a, \space b^2-c^2=n-a^2$$
and hence $n-a|n-a^2$. Since $n-a|n^2-a^2$ it follows that $n-a|n^2-n$. Since $n \ge 2$, we have $n^2-n > 0$ and thus it has only finitely many integer devisors and thus only finitely many $a$ are possible. It remains to be shown that for each such $a$, only finitely many $b,c$ can exist.
$a=n$ is impossible, as that would imply $0|n^2-n$, which is impossible. So we get 
$$b+c = \frac{b^2-c^2}{b-c} = \frac{n-a^2}{n-a}$$
Together with $b-c=n-a$ this gives exactly one solution $(b,c)$ in rationals, so at most one for integers, which concludes the proof. $\blacksquare$
These equations for $b-c$ and $b+c$ are also the way I found the special solutions given above. $n-a|n^2-n$ implies looking for divisors of $n^2-n$, where $n$ and $n-1$ obviously stand out. A little experimentation with signs lead to considering $a-n=n$ and $a-n=n-1$, and the equations for $b-c$ and $b+c$ then lead straightforward to the solutions.
A: Let's solve the system in General.
$$\left\{\!\begin{aligned}
&  a+b-c=n  \\
&  a^2+b^2-c^2=n  
\end{aligned}\right. $$
Using a replacement... solve one equation. $c= a+b-n$
$$a^2+b^2-a^2-b^2-n^2-2ab+2(a+b)n=n$$
$$2(n-b)a=n(n-2b+1)$$
$$a=\frac{n(n-2b+1)}{2(n-b)}$$
There are $4$ solutions.
$1$ decision.
$$\frac{n}{(n-b)}=t$$
$$b=n\frac{(t-1)}{t}$$ 
We will decompose on multipliers. $n=kt$
$$b=k(t-1)$$
$2$ decision.
$$\frac{n}{(n-b)}=2t$$
$$b=\frac{n(2t-1)}{2t}$$
We will decompose on multipliers. $n=kt$
$$b=\frac{k(2t-1)}{2}$$
$3$ decision.
$$\frac{(n-2b+1)}{(n-b)}=t$$
We will decompose on multipliers. $n-1=k(t-2)$
$$b=n+k$$
$4$ decision.
$$\frac{(n-2b+1)}{(n-b)}=2t$$
We will decompose on multipliers. $n-1=2k(t-1)$
$$b=n+k$$
It can be seen that the solutions are related to the factorization of the number.  Means the number of decisions of course.
